Similarities and 3D analytic geometry

On a scale plan of a district of Kinshasa, each crossroads is the image of its real position under a transformation that shrinks all distances by the same ratio: this is a similarity. In space, engineers who install a power line or a bridge describe the cables by lines and the roofs or embankments by planes. This lesson studies plane similarities, then equations of planes and lines in space.

Plane similarities

A similarity with ratio $k>0$ is a transformation of the plane that multiplies all distances by $k$: $M'N'=k\,MN$. It preserves geometric angles, collinearity and parallelism; it multiplies areas by $k^2$.

In the plane identified with $\mathbb{C}$, every direct similarity has the form $z'=az+b$ with $a\ne0$:

• its ratio is $k=|a|$ and its angle is $\theta=\arg a$;
• if $a\ne1$, it has a unique fixed point, the centre $\omega=\dfrac{b}{1-a}$, and $z'-\omega=a(z-\omega)$: it is the composition of the rotation with centre $\omega$ and angle $\theta$ and the dilation with centre $\omega$ and ratio $k$;
• if $a=1$, it is a translation.

An indirect similarity has the form $z'=a\bar z+b$: it reverses the direction of oriented angles.

Planes in space (orthonormal frame)

A plane has a Cartesian equation $ax+by+cz+d=0$, with $(a, b, c)\ne(0, 0, 0)$; the vector $\vec n(a, b, c)$ is normal (orthogonal) to it.

• Plane through $A(x_A, y_A, z_A)$ with normal vector $\vec n$: $a(x-x_A)+b(y-y_A)+c(z-z_A)=0$.
• Two planes are parallel if their normal vectors are collinear, perpendicular if their normal vectors are orthogonal.
• Distance from a point $M_0(x_0, y_0, z_0)$ to the plane: $d=\dfrac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}$.
• Plane meeting the axes at $(p, 0, 0)$, $(0, q, 0)$, $(0, 0, s)$ (nonzero): $\dfrac xp+\dfrac yq+\dfrac zs=1$.

Lines in space

The line through $A(x_A, y_A, z_A)$ with direction vector $\vec u(\alpha, \beta, \gamma)$ has the parametric representation
$$\begin{cases}x=x_A+\alpha t\\y=y_A+\beta t\\z=z_A+\gamma t\end{cases}\qquad t\in\mathbb{R}.$$
If $\alpha\beta\gamma\ne0$, we derive the Cartesian equations $\dfrac{x-x_A}{\alpha}=\dfrac{y-y_A}{\beta}=\dfrac{z-z_A}{\gamma}$; a line is also the intersection of two intersecting planes.

Line and plane: replace $x$, $y$, $z$ by their expressions in $t$ in the equation of the plane. A unique solution in $t$ gives the intersection point; no solution means the line is parallel to the plane; infinitely many mean it lies in the plane. The line is perpendicular to the plane if $\vec u$ is collinear with $\vec n$.